1. **Define the sequence $a_n$**:
For every natural number $n$, let $a_n = 20\underbrace{00\cdots0}_{n \text{ zeros}}18$. For example, $a_1 = 2018$, $a_3 = 200018$, $a_7 = 2000000018$.

2. Claim:
We claim that for each natural $n \equiv 1 \pmod{2018}$, $2018$ divides $a_n$. This is sufficient to prove the desired statement since there are infinitely many natural numbers that are congruent to $1 \pmod{2018}$.

3. **Expression for $a_n$**:
Note that $a_n$ can be expressed as:
$$a_n = 2 \cdot 10^{n+2} + 18$$

4. Using Fermat's Little Theorem (FLT):
Fermat's Little Theorem states that if $p$ is a prime and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$. Here, we use $p = 1009$ (since $2018 = 2 \times 1009$).

5. Modulo 1009:
We need to show that $2 \cdot 10^{n+2} + 18 \equiv 0 \pmod{1009}$ for $n \equiv 1 \pmod{2018}$.

Since $n \equiv 1 \pmod{2018}$, we have:
$$n = 2018k + 1 \quad \text{for some integer } k$$

6. Simplify the exponent:
$$2 \cdot 10^{n+2} + 18 \equiv 2 \cdot 10^{2018k + 3} + 18 \pmod{1009}$$

7. Using FLT:
By Fermat's Little Theorem, $10^{1008} \equiv 1 \pmod{1009}$. Therefore:
$$10^{2018k + 3} = 10^{(2 \cdot 1009k) + 3} \equiv 10^3 \pmod{1009}$$

8. **Calculate $10^3 \mod 1009$**:
$$10^3 = 1000 \equiv -9 \pmod{1009}$$

9. Substitute back:
$$2 \cdot 10^{2018k + 3} + 18 \equiv 2 \cdot (-9) + 18 \equiv -18 + 18 \equiv 0 \pmod{1009}$$

10. Combine with divisibility by 2:
Since $a_n$ ends in 8, it is always divisible by 2. Therefore, $a_n$ is divisible by both 2 and 1009. Since 2 and 1009 are relatively prime, $a_n$ is divisible by $2018$.

$\blacksquare$